Lemma 33.20.4. Let $k$ be a field. Let $f : X \to Y$ be a morphism of locally algebraic $k$-schemes.

1. For $y \in Y$, the fibre $X_ y$ is a locally algebraic scheme over $\kappa (y)$ hence all the results of Lemma 33.20.3 apply.

2. Assume $X$ is irreducible. Set $Z = \overline{f(X)}$ and $d = \dim (X) - \dim (Z)$. Then

1. $\dim _ x(X_{f(x)}) \geq d$ for all $x \in X$,

2. the set of $x \in X$ with $\dim _ x(X_{f(x)}) = d$ is dense open,

3. if $\dim (\mathcal{O}_{Z, f(x)}) \geq 1$, then $\dim _ x(X_{f(x)}) \leq d + \dim (\mathcal{O}_{Z, f(x)}) - 1$,

4. if $\dim (\mathcal{O}_{Z, f(x)}) = 1$, then $\dim _ x(X_{f(x)}) = d$,

3. For $x \in X$ with $y = f(x)$ we have $\dim _ x(X_ y) \geq \dim _ x(X) - \dim _ y(Y)$.

Proof. The morphism $f$ is locally of finite type by Morphisms, Lemma 29.15.8. Hence the base change $X_ y \to \mathop{\mathrm{Spec}}(\kappa (y))$ is locally of finite type. This proves (1). In the rest of the proof we will freely use the results of Lemma 33.20.3 for $X$, $Y$, and the fibres of $f$.

Proof of (2). Let $\eta \in X$ be the generic point and set $\xi = f(\eta )$. Then $Z = \overline{\{ \xi \} }$. Hence

$d = \dim (X) - \dim (Z) = \text{trdeg}_ k \kappa (\eta ) - \text{trdeg}_ k \kappa (\xi ) = \text{trdeg}_{\kappa (\xi )} \kappa (\eta ) = \dim _\eta (X_\xi )$

Thus parts (2)(a) and (2)(b) follow from Morphisms, Lemma 29.28.4. Parts (2)(c) and (2)(d) follow from Lemmas 33.19.3 and 33.19.1.

Proof of (3). Let $x \in X$. Let $X' \subset X$ be a irreducible component of $X$ passing through $x$ of dimension $\dim _ x(X)$. Then (2) implies that $\dim _ x(X_ y) \geq \dim (X') - \dim (Z')$ where $Z' \subset Y$ is the closure of the image of $X'$. This proves (3). $\square$

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